2017 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 2
Problem
Suppose , and all denote distinct digits from to . If , what are , and ?
Solution
and
Another way of stating the question is that = . This means that the ending digit of or has to be the ending digit of a square number, or {}. Keep in mind that is a 5-digit number. We know that cannot be 0 because would become a 2-digit number. It cannot be 4 or over because would have 6 digits instead of 5. This only leaves = 1. Because the last digit of is , we can conclude that must be or . However, since all the digits are unique, = . Now the expression becomes:
=
Because the first two digits of are and , cannot take on too many values because if it becomes , will become . It cannot be either because even then, would be too small to satisfy (If , it would still have too big a square for to be able to satisfy, being ). Now, if is tested from the set {}:
:
= (Does not work)
:
= (Does not work)
:
= (Does not work)
:
=
This works as it fits the template:
See also
2017 UNM-PNM Contest II (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
All UNM-PNM Problems and Solutions |